Show that the normal at P with x+py=2ap+ap^3 passes through point R(-ap,3a+ap^2)

Asked on by bubbletrea

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sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should remember that you may prove that a given curve passes through a given point if coordinates of the point verify the equation of curve.

Hence, you should substitute `-ap`  for x and `3a+ap^2`  for y in the given equation `x+py=2ap+ap^3`  such that:

`-ap + p(3a+ap^2) = 2ap + ap^3`

You need to open the brackets to the left side such that:

`-ap + 3ap + ap^3 = 2ap + ap^3`

Combining like terms to the left yields:

`2ap + ap^3 = 2ap + ap^3`

Hence, since the values of coordinates of the given point check the equation, hence, the normal `x+py=2ap+ap^3`  passes through the point `(-ap , 3a+ap^2).`

charithccmc's profile pic

charithccmc | (Level 2) eNoter

Posted on

x + py = 2ap + ap^3

We form this equation to ax + by + c =0

x + py - (2ap + ap^3) = 0

If above equation passes through the R point , x, y coninate of the R point must complete the above equation.

R (-ap, 3a+ap^2)

x= -ap , y= 3a+ap^2

Now we apply x, y to above equation

     x + py - (2ap + ap^3) 

     = -ap + p(3a+ap^2) - (2ap + ap^3)

     = -ap + p(3a) + p(ap^2) - (2ap + ap^3)

     = -ap + 3ap + ap^3 - 2ap - ap^3

     = 0


x + py = 2ap + ap^3 passes through R point









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